Pipes And Cistern All Competence Aptitude

Let's find the rates at which each pipe fills or empties the tank:

Pipe A fills the tank in 36 minutes, so its filling rate is \( \dfrac{1}{36} \) tank per minute.
Pipe B fills the tank in 45 minutes, so its filling rate is \( \dfrac{1}{45} \) tank per minute.
Pipe C empties the tank in 30 minutes, so its emptying rate is \( \dfrac{1}{30} \) tank per minute.

Now, let's calculate the combined rate of pipes A and B when both are opened together:

\[ \text{Combined rate of A and B} = \dfrac{1}{36} + \dfrac{1}{45} \]\[ = \dfrac{5}{180} + \dfrac{4}{180} \]\[ = \dfrac{9}{180} \]\[ = \dfrac{1}{20} \] tank per minute.

The tank starts filling with pipes A and B for 7 minutes, so the amount of tank filled in these 7 minutes is:

\[ \text{Amount filled by A and B} = \text{Rate} \times \text{Time} \]\[ = \dfrac{1}{20} \times 7 \]\[ = \dfrac{7}{20} \] of the tank.

Now, after 7 minutes, pipe C is opened, which empties the tank.

Let \( x \) be the additional time (in minutes) it takes to fill the tank after pipe C is opened.

Now, the remaining work (tank filling) is \( 1 - \dfrac{7}{20} \) since the tank was already filled with \(\dfrac{7}{20}\) of its capacity.

The combined rate of filling (pipes A and B) and emptying (pipe C) is:\[ \text{Combined rate} = \dfrac{1}{20} - \dfrac{1}{30} \]\[ = \dfrac{3}{60} - \dfrac{2}{60} \]\[ = \dfrac{1}{60} \] tank per minute.

Using the formula \( \text{Time} = \dfrac{\text{Amount of work}}{\text{Rate of work}} \), we find \( x \):

\[ x = \dfrac{1 - \dfrac{7}{20}}{\dfrac{1}{60}} \]\[ x = \dfrac{\dfrac{20}{20} - \dfrac{7}{20}}{\dfrac{1}{60}} \]\[ x = \dfrac{\dfrac{13}{20}}{\dfrac{1}{60}} \]\[ x = \dfrac{13}{20} \times 60 \]\[ x = 39 \] minutes.

Therefore, the tank is filled in \( 7 + 39 = 46 \) minutes. 

So, the tank is filled up in 46 minutes.

To find out how much time it will take to fill the cistern with 100 liters of water, let's first determine the combined rate at which the boy and the girl fill the cistern.

The boy's rate is 4 litres every 3 minutes, which means he fills \( \dfrac{4}{3} \) litres per minute.
The girl's rate is 3 litres every 4 minutes, which means she fills \( \dfrac{3}{4} \) litres per minute.

So, the combined rate of filling the cistern is \( \dfrac{4}{3} + \dfrac{3}{4} \) litres per minute.

To add these rates, let's first find a common denominator, which is 12. Then, we get:

\[ \dfrac{16}{12} + \dfrac{9}{12} = \dfrac{25}{12} \] litres per minute.

So, together, they fill \( \dfrac{25}{12} \) litres per minute.

Now, let's find out how many minutes it will take for them to fill 100 litres of water:

\[ \text{Time} = \dfrac{\text{Amount of water}}{\text{Rate}} = \dfrac{100}{\dfrac{25}{12}} \] minutes

\[ = \dfrac{100 \times 12}{25} \] minutes

\[ = \dfrac{1200}{25} \] minutes

\[ = 48 \] minutes.

So, it will take 48 minutes to fill 100 litres of water in the cistern. Therefore, the correct option is 48 minutes.

To find out how much time is needed to empty the tank when both taps operate simultaneously, we need to calculate the combined rate at which they empty the tank.

The first tap can empty the tank in 30 minutes, so its rate is \( \dfrac{1}{30} \) of the tank per minute.
The second tap can empty the tank in 45 minutes, so its rate is \( \dfrac{1}{45} \) of the tank per minute.

When both taps operate simultaneously, their combined rate is the sum of their individual rates:

\[ \dfrac{1}{30} + \dfrac{1}{45} = \dfrac{3}{90} + \dfrac{2}{90} = \dfrac{5}{90} = \dfrac{1}{18} \]

So, together, they empty \( \dfrac{1}{18} \) of the tank per minute.

To find out how much time is needed to empty the tank completely, we can use the formula:

\[ \text{Time} = \dfrac{\text{Amount of water}}{\text{Rate}} \]

Since the tank needs to be completely emptied, the amount of water is 1 (the entire tank). So, the time needed is:

\[ \text{Time} = \dfrac{1}{\dfrac{1}{18}} = 18 \] minutes.

Therefore, when both taps operate simultaneously, it takes 18 minutes to empty the tank. The correct option is 18 minutes.

Let's first find the rate at which each pipe fills the cistern.

Pipe A can fill the cistern in 6 hours, so its rate is \( \dfrac{1}{6} \) of the cistern per hour.
Pipe B can fill the cistern in 8 hours, so its rate is \( \dfrac{1}{8} \) of the cistern per hour.

When both pipes are opened simultaneously, their combined rate is the sum of their individual rates:

\[ \dfrac{1}{6} + \dfrac{1}{8} = \dfrac{4}{24} + \dfrac{3}{24} = \dfrac{7}{24} \]

So, together, they fill \( \dfrac{7}{24} \) of the cistern per hour.

When both pipes are opened simultaneously for 2 hours, they fill \( 2 \times \dfrac{7}{24} = \dfrac{14}{24} \) of the cistern.

Now, after 2 hours, only pipe B is operational. So, the remaining fraction of the cistern to be filled is \( 1 - \dfrac{14}{24} = \dfrac{10}{24} \).

Now, let's find out how many hours it will take for pipe B to fill this remaining fraction:

\[ \text{Time} = \dfrac{\text{Amount of water}}{\text{Rate}} = \dfrac{\dfrac{10}{24}}{\dfrac{1}{8}} \]

\[ = \dfrac{10}{24} \times 8 \]

\[ = \dfrac{80}{24} \]

\[ = 3 \dfrac{1}{3} \]

So, it will take \( 3 \dfrac{1}{3} \) hours for pipe B to fill the remaining part of the cistern.

Therefore, the correct answer is 3 and 1/3 hours.

Let's break down the problem step by step:

1. The first tap fills the tank in 12 hours, so its rate of filling is 1/12 tank per hour.
2. The second tap empties half the tank in 10 hours, so its rate of emptying is 1/20 tank per hour (since it empties half the tank in 10 hours, it empties 1/20 of the tank per hour).

When both taps are open, their combined effect per hour is the sum of their individual rates. So, the net rate of filling the tank when both taps are open is:

1/12 (filling rate of the first tap) - 1/20 (emptying rate of the second tap)

Now, let's find the net rate:

1/12 - 1/20 = (5/60) - (3/60) = 2/60 = 1/30

This means that when both taps are open, the tank fills at a rate of 1/30 tank per hour.

To fill half the tank, it would take double the time it takes to fill the entire tank, because if it fills half the tank, it's operating at half the rate. So, it would take 30 hours for the tank to be filled to half its capacity.

So, the correct answer is:

30 hours

Let's break down the problem step by step:

1. The first tap can fill the tank in 30 minutes, so its rate of filling is 1/30 tank per minute.
2. The second tap can fill the tank in 60 minutes, so its rate of filling is 1/60 tank per minute.

When both taps are open for 5 minutes, they will together fill a fraction of the tank:

First tap's rate * time + Second tap's rate * time = Fraction of the tank filled

(1/30 * 5) + (1/60 * 5) = (5/30) + (5/60) = (1/6) + (1/12) = (2/12) + (1/12) = 3/12 = 1/4

So, after 5 minutes, they have filled 1/4 of the tank.

Now, if only the second tap is open, it will fill the remaining 3/4 of the tank. Since the second tap fills at a rate of 1/60 tank per minute, we can find how long it will take to fill 3/4 of the tank:

Time = (Remaining fraction of tank) / (Rate of second tap)
      = (3/4) / (1/60)
      = (3/4) * (60/1)
      = 45 minutes

So, after the first tap is closed, it will take an additional 45 minutes for the tank to be full.

Therefore, the correct answer is:

45 minutes

Let's analyze each arrangement:

1. Q alone is open: Pipe Q alone fills the cistern in 8 hours.
2. P and S are open: Pipe P fills the cistern in 4 hours, and pipe S empties it in 10 hours. If both P and S are open, then the net effect is filling - emptying, which is (1/4 - 1/10) = (5/20 - 2/20) = (3/20) of the cistern per hour. So, the time taken to fill the cistern completely in this case would be 20/3 hours, which is approximately 6.67 hours.
3. P, R, and S are open: Pipes P and R fill the cistern in 4 hours and 12 hours respectively, and pipe S empties it in 10 hours. So, the net effect is (1/4 + 1/12 - 1/10) = (15/60 + 5/60 - 6/60) = (14/60) = (7/30) of the cistern per hour. Thus, the time taken to fill the cistern completely in this case would be 30/7 hours, which is approximately 4.29 hours.
4. P, Q, and S are open: Pipes P and Q fill the cistern in 4 hours and 8 hours respectively, and pipe S empties it in 10 hours. So, the net effect is (1/4 + 1/8 - 1/10) = (15/60 + 7.5/60 - 6/60) = (16.5/60) = (11/40) of the cistern per hour. Thus, the time taken to fill the cistern completely in this case would be 40/11 hours, which is approximately 3.64 hours.

Comparing the times:
1. Q alone is open: 8 hours
2. P and S are open: Approximately 6.67 hours
3. P, R, and S are open: Approximately 4.29 hours
4. P, Q, and S are open: Approximately 3.64 hours

Therefore, the arrangement that fills the empty cistern in less time than others is when pipes P, Q, and S are open.

Let's denote the rate at which the pump fills the tank as \( R_f \) (in tanks per hour) and the rate at which the leak drains the tank as \( R_d \) (in tanks per hour).

According to the problem:
1. The pump alone can fill the tank in 2 hours, so \( R_f = \dfrac{1}{2} \) tank per hour.
2. With the leak, the tank is filled in \( \dfrac{2}{3} \) more hours than usual, so the effective rate of the pump (with the leak) becomes \( \dfrac{1}{\dfrac{7}{3}} = \dfrac{3}{7} \) tank per hour.

Now, we need to find the rate at which the leak drains the tank. The difference between the normal rate of the pump and the effective rate with the leak gives us the rate at which the leak is draining the tank:

\[ R_f - R_d = \dfrac{1}{2} - R_d = \dfrac{3}{7} \]

Let's solve for \( R_d \):

\[ \dfrac{1}{2} - R_d = \dfrac{3}{7} \]\[ - R_d = \dfrac{3}{7} - \dfrac{1}{2} \]\[ - R_d = \dfrac{6}{14} - \dfrac{7}{14} \]\[ - R_d = \dfrac{-1}{14} \]\[ R_d = \dfrac{1}{14} \]

So, the leak can drain the tank at a rate of \( \dfrac{1}{14} \) tank per hour.

Now, to find out how long it would take for the leak to drain all the water off the tank, we simply take the reciprocal of \( R_d \):

\[ \text{Time taken to drain the tank} = \dfrac{1}{R_d} = \dfrac{1}{\dfrac{1}{14}} = 14 \text{ hours} \]

Therefore, the leak can drain all the water off the tank in 14 hours.